Finding the Magnetic Field Due to a Wire stretching from Infinity to 0

[B3NR4Y]

Homework Statement

An intensity current I descends down the z-axis from [itex]z = \infty[/itex] to z = 0, where
it spreads out in an isotropic way on the plane z = 0. Compute the magnetic field.

Homework Equations

The only relevant equation I can think of is Ampere's Law, [itex]\oint_\gamma \vec{B} \cdot d\vec{\ell} = \mu_0 I_{enc}[/itex]

The Attempt at a Solution

I think I should break up the problem into two parts. One to find the field due to the wire and the other to find the field due to the plane. The sum of these two will give me the total field, which is what I want.

For the wire I think the field is given by half of the field from an infinite wire. So we have that [itex]\vec{B}=\frac{\mu_0 I}{4\pi s} \hat{\phi}[/itex], but I'm not sure how this looks below the plane. I'm fine with this being the field everywhere else.

For the plane I'm not even sure how to think about it. It expands isotropically so I assume that [itex]\vec{K} = \frac{I}{r^2} \hat{r}[/itex] however I don't quite know how I will then deal with the plane to compute the field. I think I should use Ampere's law with a rectangular loop that encloses some of the above part of the plane and some of the below, but I get confused when computing it.

Thanks for any help

 

[TSny]

I think I should break up the problem into two parts. One to find the field due to the wire and the other to find the field due to the plane. The sum of these two will give me the total field, which is what I want.

For an arbitrary point P, try to determine the direction of the B field at P due to the wire and also the direction of the B field at P due to the plane.

For the wire I think the field is given by half of the field from an infinite wire.

That is not true for an arbitrary point P.

 

[Vibhor]

Hello Sir ,

For an arbitrary point P, try to determine the direction of the B field at P due to the wire and also the direction of the B field at P due to the plane.

Since OP hasn't responded , I would like to solve this problem .

I am trying to apply Ampere's law . We need to consider two regions , z>0 and z<0 .

First consider the region z>0 .

Consider a circular loop around the wire and an arbitrary point P whose coordinates are such that x>0 and y=0 . I hope there is only one such point .

The direction of magnetic field due to wire at this point will be towards -Y (negative Y direction) . Right ?

But how should I calculate direction of magnetic field due to the plane ? I am supposing that by isotropic the question means radially outwards .

Thanks .

 

[TSny]

Consider a circular loop around the wire and an arbitrary point P whose coordinates are such that x>0 and y=0 . I hope there is only one such point .

The direction of magnetic field due to wire at this point will be towards -Y (negative Y direction) . Right ?

Yes.

But how should I calculate direction of magnetic field due to the plane ? I am supposing that by isotropic the question means radially outwards .

Yes, radially outward. You can use symmetry arguments to deduce the direction.

 

[Vibhor]

You can use symmetry arguments to deduce the direction.

How ?

 

[TSny]

Pick an arbitrary point P' in the plane and consider a radial element of current at P'. Consider the B field produced at P by this current element at P'. Then pick another point P'' in the plane with a location that is "symmetrically related" to the location of P'. (You decide the location of the symmetrically related point P''.) Consider the B field at P due to a current element at P''.

Any radial current element can be broken into components parallel to the x and y-axes. You can treat these two components of the current separately.

 

[Vibhor]

Ok . I think the direction of magnetic field due to plane will be in the same direction as that of the wire . That does give the answer . I will explain my reasoning later .

Please come to the region z<0 . Now hypothetically extend the current . Again consider a circular loop . Applying Amperes' law ##\int_\vec{B} \cdot d\vec{\ell} = 0 ## .

##\vec{B} = \vec{B_{w}}+\vec{B_{p}}## . I think the direction of magnetic field due to plane would be in +Y i.e opposite to that due to wire . The net magnetic field would be tangential to the loop . So we can take B out of the integral and argue that the net magnetic field will be zero .

Is it a sensible reasoning ?

 

[TSny]

Yes. Nice.

 

[Hesch]

The problem is solved by substituting the plane by a (infinit) number of wires, symmetrical placed in the plane ( constant mutual angles ).

The current in the z-axis will form a circular field around the z-axis, and so will the wires together at some distance from the plane.

Looking in the positive direction of the current ( negative direction of the z-axis ) the magnetic field, induced by the plane, will be clockwise for z>0, and will be counter clockwise for z<0. As for the current in the z-axis, it will induce a clockwise magnetic field. This means that there must be locations for z<0, where the sum of the magnetic fields, induced by the z-axis and the plane, will be zero.

Maybe I will attach a figure later today ( monday ).

 

[Vibhor]

Pick an arbitrary point P' in the plane and consider a radial element of current at P'. Consider the B field produced at P by this current element at P'. Then pick another point P'' in the plane with a location that is "symmetrically related" to the location of P'. (You decide the location of the symmetrically related point P''.) Consider the B field at P due to a current element at P''.

Please see the attached image .

Do you think I have marked the symmetric points correctly ? The line joining them is perpendicular to Y axis .

 

[Vibhor]

The problem is solved by substituting the plane by a (infinit) number of wires, symmetrical placed in the plane ( constant mutual angles ).

I think my initial reasoning is a bit similar to how you have thought . Thanks for chipping in .

Maybe I will attach a figure later today ( monday ).

Thanks again .

 

[TSny]

Do you think I have marked the symmetric points correctly ? The line joining them is perpendicular to Y axis .

Yes, but your choice of P' and P'' looks too special since it appears that these two points have the same x-coordinate as the point P.

 

[Vibhor]

Yes, but your choice of P' and P'' looks too special since it appears that these two points have the same x-coordinate as the point P.

I think for any point on the loop , the line joining the three points will be perpendicular to the line joining the center and point P .

Right?

 

[TSny]

I think for any point on the loop , the line joining the three points will be perpendicular to the line joining the center and point P .

Right?

I'm not quite following you. Point P is located above or below the plane that carries the current. Points P' and P'' lie in the plane. So, the three points do not lie on the same straight line. For example, point P could have coordinates (x, y, z) = (5, 0, 4). Point P' might have coordinates (3, 2, 0). Then what would be the coordinates of P''?

 

[Vibhor]

Sorry . I hastily wrote something different from what I had in my mind . I will be able to reply after some time .

 

[Vibhor]

Hello Sir ,

I somehow missed following up on this problem . I came across this problem again .

For example, point P could have coordinates (x, y, z) = (5, 0, 4). Point P' might have coordinates (3, 2, 0). Then what would be the coordinates of P''?

Coordinates of P" would be (3 , -2 , 0) .

In general if O is the point where wire meets the plane and a perpendicular is dropped from P on to plane . The point where the perpendicular intersects the plane is M . For any point P' on the plane there would be a symmetric point P" on the plane such that OP' = OP" and ∠P'OM = ∠P"OM .

For example , at any point P (3,3,5) , if we consider P' to be ( 2 , 0 ,0 ) , then P" would be (0 , 2 , 0 ) .

Is that correct ?

 

[TSny]

Coordinates of P" would be (3 , -2 , 0) .

Yes

In general if O is the point where wire meets the plane and a perpendicular is dropped from P on to plane . The point where the perpendicular intersects the plane is M . For any point P' on the plane there would be a symmetric point P" on the plane such that OP' = OP" and ∠P'OM = ∠P"OM .

For example , at any point P (3,3,5) , if we consider P' to be ( 2 , 0 ,0 ) , then P" would be (0 , 2 , 0 ) .

Is that correct ?

Yes, that's right.